If the domain of the function  $\begin{array}{l}f\left(x\right)={\log }_e(4x^2+11x+6)+{\sin }^{-1}(4x+3)+{\cos }^{-1}\left(\frac{10x+6}{3}\right)\\ \end{array}$ is  $(\alpha ,\beta ],$ then $36|\alpha +\beta |$ is equal to

$$\begin{gathered} \left(i\right) 4x^2+11x+6>0 \\ x\in (-\infty ,-2)\cup (\frac{-3}{4},\infty ) \end{gathered}$$

$$\begin{gathered} \left(ii\right) 4x+3\in [-1,1] \\ x\in [-1,-1/2] \end{gathered}$$

$$\begin{gathered} \left(iii\right) \frac{10x+6}{3}\in [-1,1] \\ x\in [\frac{-9}{10},\frac{-3}{10}] \\ x\in (\frac{-3}{4},\frac{-1}{2}]\alpha =\frac{-3}{4},\beta =\frac{-1}{2} \\ \alpha +\beta =\frac{-5}{4} \\ 36|\alpha +\beta |=45 \end{gathered}$$